Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, x[10010], y[10010], isbas[10010], bas[32], S[32], b[32], Sy[10010];
vector<pair<int, int> > ans, opt;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &y[i]);
  for (int i = 1; i <= n; i++) {
    isbas[i] = -1;
    int ts = 0;
    for (int j = 30; j >= 0; j--) {
      if ((x[i] >> j) & 1) {
        if (!bas[j]) {
          isbas[i] = j;
          bas[j] = i;
          S[j] = ts | (1 << j);
          break;
        } else
          x[i] ^= x[bas[j]], ts ^= S[j];
      }
    }
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    int ts = 0;
    for (int j = 30; j >= 0; j--) {
      if ((y[i] >> j) & 1) {
        if (bas[j])
          y[i] ^= x[bas[j]], ts ^= S[j];
        else {
          puts("-1");
          return 0;
        }
      }
    }
    if (isbas[i] != -1) {
      b[cnt] = i;
      Sy[cnt++] = ts;
      continue;
    }
    ans.push_back(make_pair(i, i));
    for (int j = 30; j >= 0; j--)
      if ((ts >> j) & 1) ans.push_back(make_pair(i, bas[j]));
  }
  for (int i = 0; i < cnt; i++) {
    int ts = Sy[i];
    Sy[i] = 0;
    for (int j = 0; j < cnt; j++) {
      if ((ts >> isbas[b[j]]) & 1) Sy[i] |= (1 << j);
    }
  }
  for (int i = 0; i < cnt; i++) {
    if (!((Sy[i] >> i) & 1)) {
      for (int j = i + 1; j < cnt; j++) {
        if ((Sy[j] >> i) & 1) {
          opt.push_back(make_pair(b[i], b[j]));
          opt.push_back(make_pair(b[j], b[i]));
          opt.push_back(make_pair(b[i], b[j]));
          swap(Sy[i], Sy[j]);
          break;
        }
      }
    }
    if ((Sy[i] >> i) & 1) {
      for (int j = i + 1; j < cnt; j++)
        if ((Sy[j] >> i) & 1)
          opt.push_back(make_pair(b[j], b[i])), Sy[j] ^= Sy[i];
    }
  }
  for (int i = 0; i < cnt; i++) {
    if (!((Sy[i] >> i) & 1)) ans.push_back(make_pair(b[i], b[i]));
    for (int j = i + 1; j < cnt; j++) {
      if ((Sy[i] >> j) & 1) ans.push_back(make_pair(b[i], b[j]));
    }
  }
  reverse(opt.begin(), opt.end());
  ans.insert(ans.end(), opt.begin(), opt.end());
  printf("%d\n", ans.size());
  for (auto &x : ans) printf("%d %d\n", x.first, x.second);
  return 0;
}